Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid) what is the molar specific heat of micture of constant volume? `(R=8.3JmolK)`

Options

A. `19.7J//molK`
B. `15.7J//molK`
C. `17.4J//molK`
D. `21.6JmolK`

Answer

To find the molar specific heat at constant volume of a mixture of helium and hydrogen, we can follow these steps:

### Step 1: Identify the number of moles and degrees of freedom
We have:
- Helium (He): 2 moles
- Hydrogen (H₂): 3 moles

The degrees of freedom (f) for each gas are:
- For Helium (monatomic gas), \( f_1 = 3 \)
- For Hydrogen (diatomic gas), \( f_2 = 5 \)

### Step 2: Calculate the effective degrees of freedom for the mixture
Using the formula for the effective degrees of freedom \( f_{mix} \):
\[ f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} \]
Where:
- \( n_1 = 2 \) (moles of He)
- \( n_2 = 3 \) (moles of H₂)

Substituting the values:
\[ f_{mix} = \frac{(2 \times 3) + (3 \times 5)}{2 + 3} \]
\[ f_{mix} = \frac{6 + 15}{5} = \frac{21}{5} \]

### Step 3: Calculate the molar specific heat at constant volume
The molar specific heat at constant volume \( C_v \) can be calculated using the formula:
\[ C_v = \frac{f_{mix}}{2} R \]
Where \( R = 8.3 \, \text{J/mol K} \).

Substituting the values:
\[ C_v = \frac{21/5}{2} \times 8.3 \]
\[ C_v = \frac{21 \times 8.3}{10} = \frac{174.3}{10} = 17.43 \, \text{J/mol K} \]

### Final Answer
The molar specific heat of the mixture at constant volume is approximately \( 17.4 \, \text{J/mol K} \).

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